The starred cells mark the 56-61-17-72-25 BUG-lite in r34678c46. Applying a little multi-digit coloring on the deadly pattern results in the conjugate markings 'a' and 'A'. Note that the only cells where the deadly pattern can be avoided are r4c6 and r6c4. Note also that in those cells, the 'a' color forces you into the deadly pattern digits. Therefore, by uniqueness, color 'a' must be false and color 'A' is true. You can also represent this as
(79=25)r4c6 -BUGLT- (17=4)r6c4 - (1)r6c4 = (1-6)r7c4 = (6)r7c6 - (6=5)r3c6 => r4c6 <> 5. This takes us here...

Sorry to be slow on the chain...is it a discontinuous nice loop with the discontinuity at r4c4 due to the two (unlabeled) weak links to digit 7? And if so, how would a learner spot this loop?

If you would rather label your linkages instead of noting their endpoints, then it is the discontinuous nice loop you mention. I like the candidate-to-candidate AIC way of looking at it better.

After a cursory look for shorter xy-wing-type chains, I tend to use a multidigit multicoloring method to help find my AICs. Applying multidigit multicoloring to the relevent parts of the above grid gives you...

The candidate labels 'a' and 'A' represent a pair of conjugate colors, as do 'b' and 'B', etc. Simple multidigit coloring worked for this one as the 7 in r4c4 sees (is weakly linked to) both an 'a' and 'A', and so it can't be valid.

To show how multicoloring works, say you missed the fact that there was a 7a and 5A in the same box. We have a color AIC

A = a - B = b

Which tells us color A or b must be true. Thus we also have the 7 in r4c4 seeing both a 5A and a 7b, so it still can't be true.

[edit - mixed up my colors]

Last edited by Myth Jellies on Sat Feb 03, 2007 12:09 am; edited 1 time in total

The nodes in the chain are labelled with letters A through F in the diagram. Incidentally, in my posts I try to follow the conventions given on the Sudopedia page for Eureka notation, which differ a bit from your style. Following these conventions the chain is written as:

In this position your "27" based UR in r28c46 is very evident, but (I later found that) it's not even needed since the ALS XZ elimination represented by this grouped chain is enough to complete the solution (ALS's marked with A and B in the diagram):

Need help in understanding how inferences (and the Eureka notation) work for ALS’s and other groups. Ron used (8=593)r7c278 for the “B” nodes in r7. Those three cells form an ALS with an “extra” digit of either 3 or 8. So, to make the inferences come out right, does the Eureka notation require the first and last candidates (in the parentheses) to always be the extras, and therefore not repeated elsewhere in the set? Or is it less restrictive than that?

My own approach would have been to use single candidates and develop the equivalent chain for (8=593)r7c278 as:
(8)r7c2 = (8 – 3)r7c5 = (3)r7c8.

But, then following Ron’s lead, the 2-cell ALS in r7c78 suggests trying:
(8 = 5)r7c2 – (5 = 93)r7c78.
But, are the inferences correct in the ALS? The 3 is the extra, but the 5’s are not.

Need help in understanding how inferences (and the Eureka notation) work for ALS’s and other groups. Ron used (8=593)r7c278 for the “B” nodes in r7. Those three cells form an ALS with an “extra” digit of either 3 or 8. So, to make the inferences come out right, does the Eureka notation require the first and last candidates (in the parentheses) to always be the extras, and therefore not repeated elsewhere in the set? Or is it less restrictive than that?

The ALS digits can be ordered however you like, and the strong link symbol may be placed anywhere you like. They can all be potential extra digits. The trick is, if the digit appears in multiple places in the ALS then any weak link to that digit in the ALS must see all examples of that digit in the ALS.

It is convention to place the ALS digits being weakly linked closest to their weak link symbol.

I like placing ampersands between digits because I think...

(4=5&6&8)r158c1

...conveys better that r158c1 either contains a 4 or it contains 5 & 6 & 8 in a locked set. You could continue on with...

(4=5&6&8)r158c1 - (568=3)r4c1

or

(4=5&6&8)r158c1 - (5v6v8=3)r4c1

...where here, the 568 in r4c1 represents a choice rather than a locked set.